When 2,2-dimethylbutane is subjected to free-radical chlorination, distinct monochlorinated products are possible and of these contain asymmetric carbon atoms.

See the explanation

Explanation:

In this case, we have to remember that in the monochlorination products we only have to add one "Cl" with this in mind, we can have several options.

a) 1-chloro-2,2-dimethylbutane

b) (R)-3-chloro-2,2-dimethylbutane

c) (S)-3-chloro-2,2-dimethylbutane

d) 1-chloro-3,3-dimethylbutane

Additionally, from these 4 molecules, we will have 2 enantiomers. (R)-3-chloro-2,2-dimethylbutane and (S)-3-chloro-2,2-dimethylbutane.

See figure 1

I hope it helps!

idk

explanation:

dasdad

where's the diagram?